Difference in radius of 3rd and 4th orbit in `He^+` ion in `(R_0)_1` and in `Li^(2+)` ion is `(R_0)_2` then ratio of `(R_0)_1` TO `(R_0)_2` is

Options

A. 1:2
B. 2:1
C. 3:2
D. 2:3

Answer

To solve the problem, we need to find the difference in radius between the 3rd and 4th orbits for the `He^+` ion and the `Li^(2+)` ion, and then determine the ratio of these differences.

### Step-by-Step Solution:

1. **Understanding the Formula for Radius:**
The radius of an orbit in a hydrogen-like atom is given by the formula:
\[ r_n = \frac{n^2 h^2}{4 \pi^2 m z e^2} \]
where:
- \( n \) is the principal quantum number,
- \( z \) is the atomic number,
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( e \) is the charge of the electron.

This can be simplified to:
\[ r_n = R_0 \frac{n^2}{z} \]
where \( R_0 = \frac{h^2}{4 \pi^2 m e^2} \approx 0.529 \, \text{Å} \).

2. **Calculate the Radius for `He^+`:**
For the `He^+` ion (where \( z = 2 \)):
- For \( n = 4 \):
\[ r_4 = R_0 \frac{4^2}{2} = R_0 \frac{16}{2} = 8 R_0 \]
- For \( n = 3 \):
\[ r_3 = R_0 \frac{3^2}{2} = R_0 \frac{9}{2} = 4.5 R_0 \]
- The difference \( R_0_1 \) is:
\[ R_0_1 = r_4 - r_3 = 8 R_0 - 4.5 R_0 = 3.5 R_0 \]

3. **Calculate the Radius for `Li^(2+)`:**
For the `Li^(2+)` ion (where \( z = 3 \)):
- For \( n = 4 \):
\[ r_4 = R_0 \frac{4^2}{3} = R_0 \frac{16}{3} \]
- For \( n = 3 \):
\[ r_3 = R_0 \frac{3^2}{3} = R_0 \frac{9}{3} = 3 R_0 \]
- The difference \( R_0_2 \) is:
\[ R_0_2 = r_4 - r_3 = \left( \frac{16}{3} R_0 - 3 R_0 \right) = \left( \frac{16}{3} R_0 - \frac{9}{3} R_0 \right) = \frac{7}{3} R_0 \]

4. **Finding the Ratio \( \frac{R_0_1}{R_0_2} \):**
\[ \frac{R_0_1}{R_0_2} = \frac{3.5 R_0}{\frac{7}{3} R_0} = \frac{3.5}{\frac{7}{3}} = 3.5 \times \frac{3}{7} = \frac{10.5}{7} = \frac{3}{2} \]

### Final Answer:
The ratio of \( R_0_1 \) to \( R_0_2 \) is:
\[ \frac{R_0_1}{R_0_2} = \frac{3}{2} \]